Page 409 - Shigley's Mechanical Engineering Design

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384 Mechanical Engineering Design
Figure 7–13 y Unit load
a b j
j
The influence coefficient δ ij
x
is the deflection at i due to a i
unit load at j. x
l

Computer assistance is often used to lessen the difficulty in finding transverse deflections
of a stepped shaft. Rayleigh’s equation overestimates the critical speed.
To counter the increasing complexity of detail, we adopt a useful viewpoint.
Inasmuch as the shaft is an elastic body, we can use influence coefficients. An influence
coefficient is the transverse deflection at location i on a shaft due to a unit load at loca-
tion j on the shaft. From Table A–9–6 we obtain, for a simply supported beam with a
single unit load as shown in Fig. 7–13,

⎧
b j x i
2
2
⎪ l − b − x 2
⎪ j i x i ≤ a i
6EIl
⎪
⎨
δ ij = (7–24)
⎪ a j (l − x i ) 2 2
⎪ 2lx i − a − x x i > a i
⎪ j i
⎩ 6EIl
For three loads the influence coefficients may be displayed as
j
i 1 2 3
1 δ 11 δ 12 δ 13
2 δ 21 δ 22 δ 23
3 δ 31 δ 32 δ 33

6
Maxwell’s reciprocity theorem states that there is a symmetry about the main diago-
nal, composed of δ 11 , δ 22 , and δ 33 , of the form δ ij = δ ji . This relation reduces the work
of finding the influence coefficients. From the influence coefficients above, one can find
the deflections y 1 , y 2 , and y 3 of Eq. (7–23) as follows:

y 1 = F 1 δ 11 + F 2 δ 12 + F 3 δ 13
(7–25)
y 2 = F 1 δ 21 + F 2 δ 22 + F 3 δ 23
y 3 = F 1 δ 31 + F 2 δ 32 + F 3 δ 33
2
The forces F i can arise from weight attached w i or centrifugal forces m i ω y i . The
equation set (7–25) written with inertial forces can be displayed as
2 2 2
y 1 = m 1 ω y 1 δ 11 + m 2 ω y 2 δ 12 + m 3 ω y 3 δ 13
2 2 2
y 2 = m 1 ω y 1 δ 21 + m 2 ω y 2 δ 22 + m 3 ω y 3 δ 23
2 2 2
y 3 = m 1 ω y 1 δ 31 + m 2 ω y 2 δ 32 + m 3 ω y 3 δ 33

6 Thomson, op. cit., p. 167.